Probability Question

[DuncHen22]

I knew how to do this a while back but have forgotten:

Say you have 3 teams who all have a 60% chance of winning.

-What is the probability of at least 1 of them winning?
-How about the probability of at least 2 of them winning?

I know the probability of all 3 of them winning would be .6*.6*.6 = 21.6%, but I forgot how to calculate the "either/or" probabilities.

Thanks!

[sharpcat]

.50 to win one

.25 to win two

should be .5 x .5 = .25

[LT Profits]

Well, odds of hitting EXACTLY one = 3*(.6*.4*.4) = .288, odds of hitting EXACTLY two = 3*(.6*.6*.4) = .432 and as you said, odds of exactly three are ,216.

SO, odds of AT LEAST one = .288 + .432 + .216 = 93.6%
Odds of AT LEAST two = .432 + .216 = 64.8%

[sharpcat]

sorry I did not catch the 60% win rate

so it should be .6 x .6 = .36 or 36% of hitting 2 of them

[LT Profits]

sorry I did not catch the 60% win rate

so it should be .6 x .6 = .36 or 36% of hitting 2 of them

No, one sequence would be .6 x .6 x .4 = .144

Also, there are three possible sequences (hit-hit-miss, hit-miss-hit, miss-hit-hit), so you multiply by 3 to get 43.2% chance of EXACTLY 2 out of 3.

[DuncHen22]

Thank you LT Profits, that's exactly what I was looking for!

I knew it was something simple like that I was just completely blanking. Thanks again!